\(125\left(a+2\right)^3-1\)
\(=\left[5\left(a+2\right)\right]^3-1\)
\(=\left(5a+10\right)^3-1\)
\(=\left(5a+10-1\right)\left[\left(5a+10\right)^2-\left(5a+10\right)+1\right]\)
\(=\left(5a+10-1\right)\left[25a^2+100a+100-5a-10+1\right]\)
\(=\left(5a+9\right)\left[25a^2+95a+91\right]\)
b) \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
0,125(a+2)3-1
= (0,5)3(a+2)3-1
= (0,5)3(a+2)3-13
k bt làm nx ==
x6-1 = (x3)2-12=(x3-1)(x3+1)
\(0,125.\left(a+2\right)^3-1\)
\(=0,125.\left(a^3+3.a^2.2+3.a.2^2+2^3\right)-1\)
\(=0,125.\left(a^3+6a^2+12a+8\right)-1\)
\(=0,125a^3+0,75a^2+1,5a+1-1\)
\(=0,125a^3+0,75a^2+1,5a\)
\(x^6-1=\left(x^2\right)^3-1=\left(x^2\right)^3-1^3=\left(x^2-1\right)\left(x^4+x^2+1\right)\)
0,125(a+2)3-1
= (0,5)3(a+2)3-1
= (0,5)3(a+2)3-13
k bt làm nx ==
x6-1 = (x3)2-12=(x3-1)(x3+1)
